Problem: $f(x) = 7x^{3}+5x^{2}-4x+h(x)$ $g(x) = -2x^{2}-4x$ $h(t) = -4t-3(g(t))$ $ h(g(-3)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(-3)$ . Then we'll know what to plug into the outer function. $g(-3) = -2(-3)^{2}+(-4)(-3)$ $g(-3) = -6$ Now we know that $g(-3) = -6$ . Let's solve for $h(g(-3))$ , which is $h(-6)$ $h(-6) = (-4)(-6)-3(g(-6))$ To solve for the value of $h$ , we need to solve for the value of $g(-6)$ $g(-6) = -2(-6)^{2}+(-4)(-6)$ $g(-6) = -48$ That means $h(-6) = (-4)(-6)+(-3)(-48)$ $h(-6) = 168$